Chapter 6: Annual Cash Flow

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144
Homework
Solutions
for
Engineering
Newnan,
'--
Chapter
6: Annual
Economic
Eschenbach,
Cash
Analysis,
9th Edition
Lavelle
Flow
Analysis
1\ ~;),11~l
6 -rn"
1
f~:~ie'~J
,.,c,'
$60
~':1
$45
$30
$15
;;
c
c
c
:
,-
c
, ""
C = $15 + $15 (AlG, 10%,4)
= $15 + $15 (1.381) = ~35.72
6
!
..":::'_1.._L..
-' 2
'-:la,
i~::)I
B
B
B
B
B
j
"
,,~
$100
$100
B = [$100 + $100 (F/P, 15%,4)] (AlF, 15%,5)
= [$100 + $100 (1.749)] (0.1483) = ~40.77
6-3
I
I
$60
$45
E
E
.$30
i
j
j
E
E
--.'
E = $60 -$15 (AlG, 12%,4)
= $60 -$15 (1.359) = $39.62
I
;i'
!
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r
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Homework Solutions for Engineering Economic Analysis, 9th Edition
Newnan, Eschenbach, Lavelle
6-32
Around the Lake
$75,000
$3,000/ r
$7,500/ r
$1,500/ r
$45,000
15 ears
First Cost
Maintenance
Annual Power Loss
Pro en Taxes
Salva e Value
Useful Life
Under the Lake
$125,000
$2,000/ r
$2,500/ r
$2,500/ r
$25,000
15 ears
Around the Lake
EUAC
= $75,000 (AlP, 7%,15) + $12,000 -$45,000 (AlF, 7%,15)
= $75,000 (0.1098) + $12,000 -$45,000 (0.0398)
= $18.444
Under the Lake
EUAC
= $125,000 (AlP, 7%,15) + $7,000 -$25,000 (AlF, 7%,15)
= $125,000 (0.1098) + $7,000 -$25,000 (0.0398)
= $19.730
Go around the lake.
6-33
:'~c' ~Jji 1(..i1
Engineering Department
Estimate
$30
:
$27
~
i
'
If.;
.$9
$24
Amounts x 103
$21
$] 8
$15
$12
IOE
$6
EUAC
= $30,000 -$3,000 (AlG, 8%, 10)
= $30,000 -$3,000 (3.871)
= $18,387
;~:(r:r
t:!Yro-clean'soffer of $15.000/vr is less costlv.
"
'i
~
.
157
-160
Homework Solutions for Engineering Economic Analysis, 9th Edition
Newnan, Eschenbach,
Alternative
EUAB
i
A
-EUAC
= $845
Alternative
B
EUAB
-EUAC
To
I
Lavelle
maximize
-$3,000
= $1,400
(EUAB
(0.30672)
-$5,000
-EUAC)
= -$75.16
(0.30672)
choose
= -$133.60
alternative
A. (less
neaative
value).
6-37
Machine
X
EUAC
= $5,000
(AlP,
= $5,000
(0.2505)
8%,
5)
=$1,252
Machine
Y
EUAC
= ($8,000
=$1,106
Select
Machine
-$2,000)
(AlP,
8%,12)
+ $2,000
(0.08)
+ $150
!
..
Y.
6-38
Annual
Cost
of Diesel
Annual
Cost
of Gasoline
EUACdiesel
Fuel
= ($13,000
= [$50,000km/(35
km/i)]
x $0.48/1
= $685.71
= [$50,000km/(28
km/i)]
x $0.51/1
= $910.71
-$2,000)
+ $685.71
= $11,000
(AlP,
fuel
(0.2886)
6%,
+ $300
+ $120
4) + $2,000
repairs
+ $500
(0.06)
insurance
+ $1,485.71
= $4,780.31
EUACgasoline
= ($12,000
-$3,000)
+ $910.71
(AlP,
fuel
6%,
+ $200
3) + $3,000
repairs
+ $500
.
(0.06)
insurance
= $5,157.61
The
diesel
taxi
is more
economical.
6-39
Machine
A
EUAC
= $1,000
+ Pi
~i~-
= $1,000
+ $10,000
(AlP,
10%,4)
-$10,000
(AlF,
10%,4)
=$1,000+$1,000
,
=$2,000
(
,c..'cc
7
,
--
162
Homework Solutions for EngineeringEconomic Analysis, 9thEdition
Newnan, Eschenbach, Lavelle
6-42
I
Seven year analysis period:
Alternative A
EUAB -EUAC
= $55 -[$100
+ $100 (P/F, 10%,3)
+ $100 (P/F, 10%,6)] (AlP, 10%,7)
= $55 -[$100 + $100 (0.7513) + $100 (0.5645)] (0.2054)
= +$7.43
Alternative B
EUAB -EUAC
= $61 -[$150
+ $150 (P/F, 10%,4)] (AlP, 10%,7)
": $61 -[$150
+ $150 (0.683)] (0.2054)
=~~
.Q1!QQ'§~
.
Note: The analysis periOd is seven years, hence one cannot compare three
§:
years of A vs. four years of B, If one does, the problem is Constructed
so he will get the wrong answer.
6-4S
6-43
Us
EUACgas ": (P -S) (AlP, 1'10,n) + SL + Annual Costs
= ($2,400 -$300) (AlP, 10%,5) + $300 (0.10) + $1,200 + $300
= $2,100 (0.2638) + $30 + $1,500
= $2,084
-..
He:
Np'
EUACelectr= ($6,000 -$600) (AlP, 10%, 10) + $600 (0.10) + $750 + $50
= $5,400 (0.1627) + $60 + $800
= $1,739
..
c
NPli\
'§~l~!Jb~m~mQ-mQ!Qr
.~
6-44
.QhQQ
EUAC Comparison
EqUlv
Gravity Plan
Initial Investment: = $2.8 million (AlP, 10%, 40)
= $2.8 million (0.1023)
Annual Operation and maintenance
Annual Cost
EUAC/
, I
"' .= $286,400
~\ '
= $10,000
= $296,400
E
UACa
~
.QbQ~
)
Homework Solutions for Engineering Economic Analysis, 9th Edition
163
Newnan, Eschenbach, Lavelle
Pumping Plan
Initial Investment: = $1.4 million (AlP, 10%, 20)
= $1.4 million (0.1023)
Additional investment in 10th year:
= $200,000 (P/F, 10%, 10) (AlP, 10%, 40)
= $200,000 (0.3855) (0.1023)
= $7,890
Annual Operation and maintenance
= $25,000
Power Cost:
= $50,000 for 40 years
= $50,000
Additional Power Cost in last 30 years:
= $50,000 (F/A, 10%,30) (AlF, 10%,40)
= $50,000 (164.494) (0.00226)
= $18,590
Annual Cost
= $244,680
Select the Pumoing Plan."
= $143,200
.
6-45
Use 20 year analysis period.
I
Net Present Worth Approach
~
NPWMas. = -$250 -($250 -$10) [(P/F, 6%,4) + (P/F, 6%, 8) + (P/F, 6%, 12)
+ (P/F, 6%,16)] + $10 (P/F, 6%, 20) -$20 (PIA, 6%, 20)
= -$250 -$240 [0.7921 + 0.6274 + 0.4970 + 0.3936)
+ $10 (0.3118) -$20 (11.470)
=-$1,031
NPWBRK= -$1,000 -$10 (PIA, 6%, 20) + $100 (P/F, 6%, 20)
= -$1,000 -$10 (11.470) + $100 (0.3118)
= -$1,083
~hoose Masonite to save $52 on Present Worth of Cost.
Equivalent Uniform Annual Cost Approach
~
C,c'
EUACMas.
= $20 + $250 (AlP, 6%, 4) -$10 (AlF, 6%, 4)
= $20
+ $250
(0.2886)
-$10
!
(0.2286)
..I
'"
= $90
EUACBRK
= $10 + $1,000 (AlP, 6%, 20) -$100 (AlF, 6%, 20) .;.,,"
= $10 + $1,000 (,0.872) -$100 (0.0272)"
= $94
""!
(1:1;,
;~.,)i
"...I!J
.QhQose
Masonate to save $4 oer ~ear.
d
1-
,.r
~
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